Question: Is ${414736}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {414736}= &&{4}\cdot100000+ \\&&{1}\cdot10000+ \\&&{4}\cdot1000+ \\&&{7}\cdot100+ \\&&{3}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {414736}= &&{4}(99999+1)+ \\&&{1}(9999+1)+ \\&&{4}(999+1)+ \\&&{7}(99+1)+ \\&&{3}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {414736}= &&\gray{4\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {4}+{1}+{4}+{7}+{3}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${414736}$ is divisible by $3$ if ${ 4}+{1}+{4}+{7}+{3}+{6}$ is divisible by $3$ Add the digits of ${414736}$ $ {4}+{1}+{4}+{7}+{3}+{6} = {25} $ If ${25}$ is divisible by $3$ , then ${414736}$ must also be divisible by $3$ ${25}$ is not divisible by $3$, therefore ${414736}$ must not be divisible by $3$.